# How do you differentiate f(x)=x/(1+sqrtx)?

Nov 25, 2016

f'(x)= (2+sqrtx)/(2(1+sqrtx)^2

#### Explanation:

$f \left(x\right) \text{ }$ is differentiated by applying the quotient rule
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differentiation.
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Quotient Rule:
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color(blue)((u/v)'=(u'v - v'u)/v^2
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$f ' \left(x\right) = \left(\frac{x}{1 + \sqrt{x}}\right) '$
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$f ' \left(x\right) = \textcolor{b l u e}{\frac{x ' \left(1 + \sqrt{x}\right) - \left(1 + \sqrt{x}\right) ' x}{1 + \sqrt{x}} ^ 2}$
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$f ' \left(x\right) = \frac{1 \left(1 + \sqrt{x}\right) - \left(0 + \frac{1}{2 \sqrt{x}}\right) x}{1 + \sqrt{x}} ^ 2$
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$f ' \left(x\right) = \frac{1 + \sqrt{x} - \left(\frac{1}{2 \sqrt{x}}\right) x}{1 + \sqrt{x}} ^ 2$
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$f ' \left(x\right) = \frac{1 + \sqrt{x} - \frac{x}{2 \sqrt{x}}}{1 + \sqrt{x}} ^ 2$
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$f ' \left(x\right) = \frac{\frac{2 \sqrt{x} + 2 x - x}{2 \sqrt{x}}}{1 + \sqrt{x}} ^ 2$
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$f ' \left(x\right) = \frac{\frac{2 \sqrt{x} + x}{2 \sqrt{x}}}{1 + \sqrt{x}} ^ 2$
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$f ' \left(x\right) = \frac{\frac{\sqrt{x} \left(2 + \sqrt{x}\right)}{2 \sqrt{x}}}{1 + \sqrt{x}} ^ 2$
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$f ' \left(x\right) = \frac{\frac{2 + \sqrt{x}}{2}}{1 + \sqrt{x}} ^ 2$
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f'(x)= (2+sqrtx)/(2(1+sqrtx)^2